TS EAMCET 20-July-2022 Shift 1 Solved Paper

In an isobaric process,
∆Q=∆U+∆W....(i)
Here, ∆Q=700J
Gas is given diatomic so, Cp=‌

7

2

R and CV=‌

5

2

R
Heat absorbed by gas is
∆Q‌=nCp∆T
700‌=n(‌

7

2

R)⋅∆T
⇒‌‌n∆T‌=‌

700×2

7R

=‌

200

R

...(ii)
So, change in internal energy of gas,
∆U‌=nCV∆T=CV.n∆T
‌=‌

5

2

R×‌

200

R

=500J
[from Eq. (ii)]
So, from Eq. (i), work done is
∆W=∆Q−∆U=700−500=200J